Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
template <class C, class E>
inline bool contains(const C& container, const E& element) {
  return container.find(element) != container.end();
}
template <class T>
inline void checkmin(T& a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T& a, T b) {
  if (b > a) a = b;
}
using namespace std;
vector<pair<long long, long long> > obs;
vector<long long> sombrasSum;
int sy, aa, bb, n;
double Suma(int from, int to) {
  if (from > to) return 0;
  if (from >= n) return 0;
  if (to < 0) return 0;
  return sombrasSum[to + 1] - sombrasSum[from];
}
double ProyectFenceToLuz(pair<long long, long long> p, long long fx) {
  return (double)fx + (double)sy * (double)(p.first - fx) / (double)p.second;
}
double ProyectLuzToFence(pair<long long, long long> p, long long l) {
  return (double)l -
         (double)sy * (double)(p.first - l) / ((double)p.second - (double)sy);
}
double Process(pair<long long, long long> f, pair<long long, long long> p) {
  double s = ProyectFenceToLuz(p, f.first);
  double e = ProyectFenceToLuz(p, f.second);
  checkmin(s, (double)bb);
  checkmax(s, (double)aa);
  checkmin(e, (double)bb);
  checkmax(e, (double)aa);
  return e - s;
}
void Solve(pair<long long, long long> p) {
  double apd = ProyectLuzToFence(p, aa) - 1;
  double bpd = ProyectLuzToFence(p, bb) + 1;
  checkmin(apd, 1000000000.0);
  checkmin(bpd, 1000000000.0);
  checkmax(apd, 0.0);
  checkmax(bpd, 0.0);
  long long ap = apd, bp = bpd;
  int fa = upper_bound((obs).begin(), (obs).end(), make_pair(ap, -1LL)) -
           obs.begin();
  int fb = upper_bound((obs).begin(), (obs).end(), make_pair(bp, -1LL)) -
           obs.begin();
  if (fa != 0) --fa;
  if (fb == n) --fb;
  double res = 0;
  for (int i = (0), _b = ((3) - 1); i <= _b; i++) {
    if (fa < n && fa <= fb) {
      res += Process(obs[fa], p);
      ++fa;
    }
  }
  for (int i = (0), _b = ((3) - 1); i <= _b; i++) {
    if (fb >= 0 && fb < n && fb >= fa) {
      res += Process(obs[fb], p);
      --fb;
    }
  }
  double scale = (double)(-sy + p.second) / (double)p.second;
  cout << res + Suma(fa, fb) * scale << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << std::setprecision(15);
  cout << std::fixed;
  cin >> sy >> aa >> bb >> n;
  obs.resize(n);
  sombrasSum.resize(n + 1);
  for (int i = (0), _b = ((n)-1); i <= _b; i++) {
    cin >> obs[i].first >> obs[i].second;
    sombrasSum[i + 1] = sombrasSum[i] + (obs[i].second - obs[i].first);
  }
  int q;
  cin >> q;
  for (int i = (0), _b = ((q)-1); i <= _b; i++) {
    pair<long long, long long> p;
    cin >> p.first >> p.second;
    Solve(p);
  }
  return 0;
}